ISEE Upper Level Quiz: Single And Compound Probability
20 questions · exam conditions
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Single And Compound ProbabilityQuestion 1 of 20

In a certain school, 60% of students play sports, 40% play music, and 15% play both sports and music. If a student is selected at random, what is the probability that the student plays sports given that the student plays music?

38\frac{3}{8}
1540\frac{15}{40}
14\frac{1}{4}
1560\frac{15}{60}
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ISEE Upper Level Quiz

ISEE Upper Level Quiz: Single And Compound Probability

Practice Single And Compound Probability in ISEE Upper Level with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Single And Compound Probability, giving you a quick way to practice the rules, question types, and explanations that matter most for ISEE Upper Level.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

In a certain school, 60% of students play sports, 40% play music, and 15% play both sports and music. If a student is selected at random, what is the probability that the student plays sports given that the student plays music?

  1. 38\frac{3}{8} (correct answer)
  2. 1540\frac{15}{40}
  3. 14\frac{1}{4}
  4. 1560\frac{15}{60}
Explanation: When you see "given that" in a probability question, you're dealing with conditional probability. This means you're finding the probability of one event happening when you already know another event has occurred. The question asks for P(plays sports | plays music), which reads as "the probability a student plays sports given that the student plays music." To find this, you use the conditional probability formula: P(A|B) = P(A and B) ÷ P(B). Here, you need P(sports and music) ÷ P(music) = 15% ÷ 40% = 1540=38\frac{15}{40} = \frac{3}{8}. Think of it this way: among the 40% of students who play music, 15% play both sports and music. So 1540\frac{15}{40} of music students also play sports. Looking at the wrong answers: Choice B shows 1540\frac{15}{40}, which is the correct calculation before simplifying to lowest terms. While mathematically equivalent to 38\frac{3}{8}, choice A is in simplest form. Choice C gives 14\frac{1}{4}, which you might get by incorrectly using 1560\frac{15}{60} (the percentage who play both divided by those who play sports). Choice D shows 1560\frac{15}{60} exactly, representing the common error of finding P(music | sports) instead of P(sports | music). The key strategy is to carefully identify what's given and what you're looking for. "Given that" tells you the denominator in conditional probability - it's your restricted sample space. Always double-check that you have the conditional relationship in the right direction.

Question 2

Lottery Draw: Numbers 1 through 10 are used, and 3 numbers are drawn without replacement. A player chooses 3 numbers, and the order does not matter. What is the probability of matching exactly 2 of the 3 drawn numbers?

  1. 740\frac{7}{40} (correct answer)
  2. 1120\frac{1}{120}
  3. 310\frac{3}{10}
  4. 18\frac{1}{8}
Explanation: This question tests upper-level ISEE mathematics skills: calculating probability of single and compound events. Probability is the measure of the likelihood that an event will occur, calculated by dividing the number of favorable outcomes by the total number of possible outcomes. In this scenario, 3 numbers are drawn from 1 to 10, and a player chooses 3, finding probability of exactly 2 matches (order irrelevant). Choice A is correct because there are C(3,2) × C(7,1) = 21 favorable combinations out of C(10,3) = 120, yielding 7/40. Choice B is incorrect due to using full match instead, demonstrating a common error where students confuse exact with full matches. To help students: Teach them to carefully analyze whether events are independent or dependent, and practice calculating probabilities using real-life scenarios. Encourage the use of probability trees or diagrams to visualize complex problems and identify potential errors.

Question 3

Card Draw: A 52-card deck has 13 hearts and 39 non-hearts. A student draws one card, then a second card without replacement. If the first card is a heart, what is the probability the second card is also a heart?

  1. 14\frac{1}{4}
  2. 1251\frac{12}{51} (correct answer)
  3. 1352\frac{13}{52}
  4. 1252\frac{12}{52}
Explanation: This question tests upper-level ISEE mathematics skills: calculating probability of single and compound events. Probability is the measure of the likelihood that an event will occur, calculated by dividing the number of favorable outcomes by the total number of possible outcomes. In this scenario, after drawing a heart first without replacement, students must find the conditional probability of drawing another heart. Choice B is correct because 12 hearts remain out of 51 cards, yielding 12/51. Choice A is incorrect due to using the unconditional probability, demonstrating a common error where students ignore the given condition. To help students: Teach them to carefully analyze whether events are independent or dependent, and practice calculating probabilities using real-life scenarios. Encourage the use of probability trees or diagrams to visualize complex problems and identify potential errors.

Question 4

In a dice game, you roll two fair six-sided dice, so there are 36 equally likely outcomes. The event DD is "you roll doubles," which includes (1,1)(1,1) through (6,6)(6,6). What is the probability P(D)P(D)?

  1. 112\frac{1}{12}
  2. 16\frac{1}{6} (correct answer)
  3. 536\frac{5}{36}
  4. 118\frac{1}{18}
Explanation: This question tests upper-level ISEE mathematics skills: calculating probability of single and compound events. Probability is the measure of the likelihood that an event will occur, calculated by dividing the number of favorable outcomes by the total number of possible outcomes. In this scenario, students must identify all doubles when rolling two dice: (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6). Choice B is correct because there are exactly 6 doubles out of 36 total outcomes, giving us 6/36 = 1/6. Choice A incorrectly suggests 3 doubles (1/12 = 3/36), while choice C overcounts with 5/36. To help students: Create a visual representation of all 36 outcomes and highlight the diagonal where both dice show the same number. Emphasize that doubles form a pattern along the main diagonal of the outcome grid.

Question 5

Weather Forecast: A city records 100 summer days. The data show these outcomes: 30 days have rain, 40 days have high temperatures, and 12 days have both rain and high temperatures. Calculate the probability of both rain and high temperatures on a randomly chosen day.

  1. 12%12\% (correct answer)
  2. 18%18\%
  3. 48%48\%
  4. 28%28\%
Explanation: This question tests upper-level ISEE mathematics skills: calculating probability of single and compound events. Probability is the measure of the likelihood that an event will occur, calculated by dividing the number of favorable outcomes by the total number of possible outcomes. In this scenario, weather data over 100 days is given, and students must find the probability of both rain and high temperatures. Choice A is correct because 12 days have both out of 100, yielding 12%. Choice B is incorrect due to adding instead of finding intersection, demonstrating a common error where students confuse union with intersection. To help students: Teach them to carefully analyze whether events are independent or dependent, and practice calculating probabilities using real-life scenarios. Encourage the use of probability trees or diagrams to visualize complex problems and identify potential errors.

Question 6

Card Draw: A standard 52-card deck has these outcomes: 13 hearts, 13 spades, 13 diamonds, and 13 clubs. A student draws one card, then a second card without replacement. Calculate the probability of drawing a heart and then a spade.

  1. 116\frac{1}{16}
  2. 13204\frac{13}{204} (correct answer)
  3. 14\frac{1}{4}
  4. 18\frac{1}{8}
Explanation: This question tests upper-level ISEE mathematics skills: calculating probability of single and compound events. Probability is the measure of the likelihood that an event will occur, calculated by dividing the number of favorable outcomes by the total number of possible outcomes. In this scenario, two cards are drawn without replacement from a 52-card deck, and students must find the probability of a heart followed by a spade. Choice B is correct because the probability is (13/52) × (13/51), simplifying to 13/204 after accounting for the dependent events. Choice A is incorrect due to treating the draws as independent, demonstrating a common error where students ignore the without-replacement condition. To help students: Teach them to carefully analyze whether events are independent or dependent, and practice calculating probabilities using real-life scenarios. Encourage the use of probability trees or diagrams to visualize complex problems and identify potential errors.

Question 7

Card Draw: A 52-card deck contains 13 hearts and 13 spades. A student draws two cards without replacement. What is the probability of drawing a heart and then a spade, if the student incorrectly treats the events as independent?

  1. 13204\frac{13}{204}
  2. 18\frac{1}{8}
  3. 116\frac{1}{16} (correct answer)
  4. 317\frac{3}{17}
Explanation: This question tests upper-level ISEE mathematics skills: calculating probability of single and compound events. Probability is the measure of the likelihood that an event will occur, calculated by dividing the number of favorable outcomes by the total number of possible outcomes. In this scenario, two cards are drawn without replacement, but the student assumes independence, and students must find the miscalculated probability of heart then spade. Choice C is correct because under independence assumption, it's (13/52) × (13/52) = 1/16. Choice A is incorrect due to confusing it with the actual dependent probability, demonstrating a common error where students mix up calculation methods. To help students: Teach them to carefully analyze whether events are independent or dependent, and practice calculating probabilities using real-life scenarios. Encourage the use of probability trees or diagrams to visualize complex problems and identify potential errors.

Question 8

Card Draw: A 52-card deck has 13 hearts and 39 non-hearts. A student draws two cards without replacement. What is the probability of drawing at least one heart?

  1. 12\frac{1}{2}
  2. 1534\frac{15}{34} (correct answer)
  3. 1352\frac{13}{52}
  4. 3952\frac{39}{52}
Explanation: This question tests upper-level ISEE mathematics skills: calculating probability of single and compound events. Probability is the measure of the likelihood that an event will occur, calculated by dividing the number of favorable outcomes by the total number of possible outcomes. In this scenario, two cards are drawn without replacement, and students must find the probability of at least one heart. Choice B is correct because it's 1 - (39/52 × 38/51) = 15/34 after calculating the complement. Choice A is incorrect due to oversimplifying to half, demonstrating a common error where students ignore dependencies. To help students: Teach them to carefully analyze whether events are independent or dependent, and practice calculating probabilities using real-life scenarios. Encourage the use of probability trees or diagrams to visualize complex problems and identify potential errors.

Question 9

Lottery Draw: Numbers 1 through 10 are used, and 3 numbers are drawn without replacement. The single event EE is "the first number drawn matches one chosen number." What is the probability P(E)P(E) if a player chooses 3 different numbers?

  1. 310\frac{3}{10} (correct answer)
  2. 1120\frac{1}{120}
  3. 710\frac{7}{10}
  4. 130\frac{1}{30}
Explanation: This question tests upper-level ISEE mathematics skills: calculating probability of single and compound events. Probability is the measure of the likelihood that an event will occur, calculated by dividing the number of favorable outcomes by the total number of possible outcomes. In this scenario, 3 numbers are drawn from 1 to 10, and students must find the probability that the first drawn matches one of the player's 3 chosen numbers. Choice A is correct because there are 3 favorable out of 10 possible for the first draw, yielding 3/10. Choice B is incorrect due to confusing with the full match probability, demonstrating a common error where students overcomplicate single events. To help students: Teach them to carefully analyze whether events are independent or dependent, and practice calculating probabilities using real-life scenarios. Encourage the use of probability trees or diagrams to visualize complex problems and identify potential errors.

Question 10

Lottery Draw: A player chooses 3 numbers from 1 through 10, and 3 numbers are drawn without replacement. Which is more likely: matching the first drawn number, or winning by matching all 3 numbers?

  1. Winning by matching all 3 numbers is more likely.
  2. Matching the first drawn number is more likely. (correct answer)
  3. They are equally likely.
  4. Neither event has a probability that can be found.
Explanation: This question tests upper-level ISEE mathematics skills: calculating probability of single and compound events. Probability is the measure of the likelihood that an event will occur, calculated by dividing the number of favorable outcomes by the total number of possible outcomes. In this scenario, a player chooses 3 numbers from 1 to 10, and 3 are drawn, comparing probability of matching the first versus all 3. Choice B is correct because matching the first is 3/10, far higher than 1/120 for all 3. Choice A is incorrect due to reversing the comparison, demonstrating a common error where students underestimate compound event rarity. To help students: Teach them to carefully analyze whether events are independent or dependent, and practice calculating probabilities using real-life scenarios. Encourage the use of probability trees or diagrams to visualize complex problems and identify potential errors.

Question 11

Weather Forecast: In 100 days, high temperatures happen on 40 days, and both rain and high temperatures happen on 12 days. If temperatures are high, what is the probability that it rains?

  1. 1240\frac{12}{40} (correct answer)
  2. 1230\frac{12}{30}
  3. 30100\frac{30}{100}
  4. 40100\frac{40}{100}
Explanation: This question tests upper-level ISEE mathematics skills: calculating probability of single and compound events. Probability is the measure of the likelihood that an event will occur, calculated by dividing the number of favorable outcomes by the total number of possible outcomes. In this scenario, given weather data over 100 days, students must find the conditional probability of rain given high temperatures. Choice A is correct because it's 12 days both divided by 40 hot days, yielding 12/40. Choice B is incorrect due to reversing the conditioning events, demonstrating a common error where students mix up P(A|B) with P(B|A). To help students: Teach them to carefully analyze whether events are independent or dependent, and practice calculating probabilities using real-life scenarios. Encourage the use of probability trees or diagrams to visualize complex problems and identify potential errors.

Question 12

A weather report says P(R)=40%P(R)=40\% for rain and P(H)=30%P(H)=30\% for high temperatures. The report treats RR and HH as independent events. What is the likelihood of both rain and high temperatures?

  1. 70%70\%
  2. 12%12\% (correct answer)
  3. 28%28\%
  4. 10%10\%
Explanation: This question tests upper-level ISEE mathematics skills: calculating probability of single and compound events. Probability is the measure of the likelihood that an event will occur, calculated by dividing the number of favorable outcomes by the total number of possible outcomes. In this scenario with independent weather events, students must multiply the individual probabilities: P(R and H) = P(R) × P(H) = 0.40 × 0.30. Choice B is correct because 0.40 × 0.30 = 0.12 = 12%. Choice A incorrectly adds the probabilities (40% + 30% = 70%) instead of multiplying them. To help students: Remember that 'and' means multiply for independent events, while 'or' involves addition with adjustment for overlap. Convert percentages to decimals before multiplying, then convert back to percentage form.

Question 13

In a card draw, you draw two cards from a 52-card deck without replacement. The event HH is "first card is a heart." If the first card is a heart, what is the probability the second card is a spade?

  1. 1351\frac{13}{51} (correct answer)
  2. 14\frac{1}{4}
  3. 1251\frac{12}{51}
  4. 1352\frac{13}{52}
Explanation: This question tests upper-level ISEE mathematics skills: calculating probability of single and compound events. Probability is the measure of the likelihood that an event will occur, calculated by dividing the number of favorable outcomes by the total number of possible outcomes. In this scenario, given that the first card is a heart, students must find the conditional probability that the second card is a spade from the remaining 51 cards. Choice A is correct because all 13 spades remain in the deck of 51 cards, giving probability 13/51. Choice B incorrectly uses 1/4, which would be the probability if we replaced the first card. To help students: Understand that conditional probability changes the sample space. Practice problems where the first event affects what's available for the second event.

Question 14

Two dice are rolled simultaneously. What is the probability that their sum is greater than 8 OR at least one die shows a 6?

  1. 512\frac{5}{12}
  2. 712\frac{7}{12}
  3. 23\frac{2}{3} (correct answer)
  4. 34\frac{3}{4}
Explanation: When you encounter probability questions with "OR" conditions, you're dealing with the union of events. The key insight is recognizing when events overlap and using the inclusion-exclusion principle: P(A OR B) = P(A) + P(B) - P(A AND B). Let's define our events: A = sum greater than 8, and B = at least one die shows 6. First, find P(A). The favorable outcomes for sums greater than 8 are: (3,6), (4,5), (4,6), (5,4), (5,5), (5,6), (6,3), (6,4), (6,5), (6,6) — that's 10 outcomes out of 36 possible, so P(A) = 1036\frac{10}{36}. Next, find P(B). It's easier to calculate the complement: P(no 6's) = 56×56=2536\frac{5}{6} \times \frac{5}{6} = \frac{25}{36}, so P(at least one 6) = 1−2536=11361 - \frac{25}{36} = \frac{11}{36}. Now find P(A AND B) — outcomes where sum > 8 AND at least one die shows 6: (3,6), (4,6), (5,6), (6,3), (6,4), (6,5), (6,6). That's 7 outcomes, so P(A AND B) = 736\frac{7}{36}. Therefore: P(A OR B) = 1036+1136−736=1436=23\frac{10}{36} + \frac{11}{36} - \frac{7}{36} = \frac{14}{36} = \frac{2}{3}, which is answer C. Answer A (512\frac{5}{12}) likely comes from forgetting to subtract the overlap. Answer B (712\frac{7}{12}) might result from miscounting favorable outcomes. Answer D (34\frac{3}{4}) could come from incorrectly adding probabilities without considering overlap. Remember: always subtract the intersection when calculating "OR" probabilities to avoid double-counting overlapping outcomes.

Question 15

A spinner has three sections: red (probability 0.4), blue (probability 0.3), and green (probability 0.3). The spinner is spun twice. What is the probability that red appears exactly once?

  1. 0.32
  2. 0.48 (correct answer)
  3. 0.52
  4. 0.64
Explanation: When you encounter probability questions involving "exactly" a certain number of outcomes over multiple trials, you're dealing with binomial probability. The key insight is that "exactly once" means the event happens in one trial but not the other. For red to appear exactly once in two spins, there are two possible scenarios: red on the first spin and not red on the second, OR not red on the first spin and red on the second. Since these scenarios are mutually exclusive, you add their probabilities. First scenario: Red first (probability 0.4), then not red (probability 0.6, since blue and green together total 0.3 + 0.3 = 0.6). This gives us 0.4×0.6=0.240.4 \times 0.6 = 0.24. Second scenario: Not red first (probability 0.6), then red (probability 0.4). This gives us 0.6×0.4=0.240.6 \times 0.4 = 0.24. Total probability: 0.24+0.24=0.480.24 + 0.24 = 0.48, which is answer choice B. Looking at the wrong answers: A (0.32) likely comes from calculating just one scenario instead of both. C (0.52) might result from incorrectly adding 0.4 + 0.3 + 0.3 - 0.4 or similar confusion about complement probabilities. D (0.64) appears to be the probability of getting red at least once, calculated as 1−(0.6)2=1−0.36=0.641 - (0.6)^2 = 1 - 0.36 = 0.64. Remember: For "exactly" problems in binomial situations, always consider all the ways the specified outcome can occur, calculate each scenario's probability, then add them together.

Question 16

Two events A and B are independent with P(A) = 0.6 and P(B) = 0.4. What is P(A or B but not both)?

  1. 0.24
  2. 0.52 (correct answer)
  3. 0.76
  4. 1.00
Explanation: When you encounter probability questions involving "A or B but not both," you're dealing with the exclusive or (XOR) - events that can happen individually but not simultaneously. Since events A and B are independent, you can find P(A or B but not both) by calculating P(A and not B) + P(not A and B). For independent events, P(A and not B) = P(A) × P(not B) = 0.6 × (1 - 0.4) = 0.6 × 0.6 = 0.36. Similarly, P(not A and B) = P(not A) × P(B) = (1 - 0.6) × 0.4 = 0.4 × 0.4 = 0.16. Therefore, P(A or B but not both) = 0.36 + 0.16 = 0.52. Choice A (0.24) represents P(A and B) = 0.6 × 0.4 = 0.24, which is the probability that both events occur together - the opposite of what we want. Choice C (0.76) equals P(A or B) = P(A) + P(B) - P(A and B) = 0.6 + 0.4 - 0.24 = 1.00 - 0.24 = 0.76, which includes cases where both events happen. Choice D (1.00) would mean certainty, which only occurs if the events were mutually exclusive and collectively exhaustive. Remember this pattern: "A or B but not both" always equals P(A or B) - P(A and B). This formula works whether events are independent or not, making it a reliable approach for exclusive or problems.

Question 17

A fair six-sided die is rolled four times. What is the probability of getting exactly three even numbers?

  1. 14\frac{1}{4} (correct answer)
  2. 416\frac{4}{16}
  3. 1264\frac{12}{64}
  4. 2764\frac{27}{64}
Explanation: This is a binomial probability problem where you need to find the probability of getting exactly three successes (even numbers) in four independent trials (die rolls). First, identify the probability of success on each roll. A six-sided die has three even numbers (2, 4, 6) out of six possible outcomes, so P(even)=36=12P(\text{even}) = \frac{3}{6} = \frac{1}{2}. The probability of failure is P(odd)=12P(\text{odd}) = \frac{1}{2}. Use the binomial probability formula: P(X=k)=(nk)⋅pk⋅(1−p)n−kP(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}, where n=4n = 4 rolls, k=3k = 3 successes, and p=12p = \frac{1}{2}. Calculate: P(X=3)=(43)⋅(12)3⋅(12)1=4⋅18⋅12=4⋅116=416=14P(X = 3) = \binom{4}{3} \cdot \left(\frac{1}{2}\right)^3 \cdot \left(\frac{1}{2}\right)^1 = 4 \cdot \frac{1}{8} \cdot \frac{1}{2} = 4 \cdot \frac{1}{16} = \frac{4}{16} = \frac{1}{4} Choice A gives 14\frac{1}{4}, which is correct. Choice B shows 416\frac{4}{16}, which equals 14\frac{1}{4} but wasn't simplified—this represents the same correct answer in unreduced form. Choice C shows 1264\frac{12}{64}, which simplifies to 316\frac{3}{16}—this likely comes from miscalculating the combinations or using wrong probability values. Choice D shows 2764\frac{27}{64}, which doesn't match any reasonable calculation for this scenario and may result from confusion with other probability formulas. When solving binomial probability problems, always identify your success probability first, then carefully apply the formula. Remember that (43)=4\binom{4}{3} = 4, representing the four different ways to arrange three successes in four trials.

Question 18

A bag contains 5 red, 4 blue, and 6 yellow marbles. Two marbles are drawn without replacement. What is the probability that they are different colors?

  1. 74105\frac{74}{105} (correct answer)
  2. 31105\frac{31}{105}
  3. 62105\frac{62}{105}
  4. 43105\frac{43}{105}
Explanation: When you encounter probability questions involving "without replacement," you need to consider how each draw affects the subsequent outcomes. This question asks for the probability that two marbles are different colors, which means avoiding the case where both marbles are the same color. The bag contains 15 total marbles (5 red + 4 blue + 6 yellow). To find the probability of different colors, calculate 1 minus the probability of same colors. The probability of drawing two marbles of the same color equals:
  • P(both red) = 515×414=20210\frac{5}{15} \times \frac{4}{14} = \frac{20}{210}
  • P(both blue) = 415×314=12210\frac{4}{15} \times \frac{3}{14} = \frac{12}{210}
  • P(both yellow) = 615×514=30210\frac{6}{15} \times \frac{5}{14} = \frac{30}{210}
Adding these: 20+12+30210=62210=31105\frac{20 + 12 + 30}{210} = \frac{62}{210} = \frac{31}{105} Therefore, P(different colors) = 1−31105=741051 - \frac{31}{105} = \frac{74}{105}, which is answer A. Answer B (31105\frac{31}{105}) represents the probability of same colors—the complement of what we want. Answer C (62105\frac{62}{105}) is the unreduced fraction 62210\frac{62}{210} incorrectly placed over 105, showing a computational error. Answer D (43105\frac{43}{105}) likely results from miscounting marble combinations or calculation mistakes. Remember: when probability questions ask for "different" or "at least one," often the complement approach (1 minus the unwanted outcome) is more efficient than calculating all desired cases separately.

Question 19

In a game with three rounds, the probability of winning each round is 0.7. What is the probability of winning at least two rounds?

  1. 0.441
  2. 0.784 (correct answer)
  3. 0.832
  4. 0.973
Explanation: When you encounter probability questions asking for "at least" a certain number of successes, you're dealing with binomial probability. The key insight is that "at least two rounds" means exactly two rounds OR exactly three rounds. To find the probability of winning exactly k rounds out of 3, use the binomial formula: P(X=k)=(nk)⋅pk⋅(1−p)n−kP(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}, where n = 3 rounds and p = 0.7. For exactly 2 wins: P(X=2)=(32)⋅(0.7)2⋅(0.3)1=3⋅0.49⋅0.3=0.441P(X = 2) = \binom{3}{2} \cdot (0.7)^2 \cdot (0.3)^1 = 3 \cdot 0.49 \cdot 0.3 = 0.441 For exactly 3 wins: P(X=3)=(33)⋅(0.7)3⋅(0.3)0=1⋅0.343⋅1=0.343P(X = 3) = \binom{3}{3} \cdot (0.7)^3 \cdot (0.3)^0 = 1 \cdot 0.343 \cdot 1 = 0.343 Therefore, P(at least 2 wins) = 0.441 + 0.343 = 0.784. Choice A (0.441) represents only the probability of winning exactly two rounds—this is a common trap where students forget to include the "exactly three" case. Choice C (0.832) likely comes from calculation errors in the binomial coefficients or powers. Choice D (0.973) is far too high and might result from misunderstanding what "at least two" means or incorrectly calculating 1 minus the probability of winning zero or one round. Remember: when you see "at least" in probability questions, always break it down into separate "exactly" cases and add them together. This systematic approach prevents missing scenarios.

Question 20

A student takes a 10-question true/false test by guessing. What is the probability that the student gets more than 7 questions correct?

  1. 7128\frac{7}{128}
  2. 11128\frac{11}{128}
  3. 561024\frac{56}{1024} (correct answer)
  4. 1761024\frac{176}{1024}
Explanation: When you encounter a question about repeated independent trials with only two outcomes (like true/false), you're dealing with binomial probability. Each guess has a 12\frac{1}{2} chance of being correct, and you need to find the probability of getting "more than 7" correct, which means exactly 8, 9, or 10 correct answers. The binomial probability formula is P(X=k)=(nk)⋅pk⋅(1−p)n−kP(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}, where n=10n = 10, p=12p = \frac{1}{2}, and kk is the number of correct answers. For 8 correct: P(X=8)=(108)⋅(12)10=45⋅11024=451024P(X = 8) = \binom{10}{8} \cdot \left(\frac{1}{2}\right)^{10} = 45 \cdot \frac{1}{1024} = \frac{45}{1024} For 9 correct: P(X=9)=(109)⋅(12)10=10⋅11024=101024P(X = 9) = \binom{10}{9} \cdot \left(\frac{1}{2}\right)^{10} = 10 \cdot \frac{1}{1024} = \frac{10}{1024} For 10 correct: P(X=10)=(1010)⋅(12)10=1⋅11024=11024P(X = 10) = \binom{10}{10} \cdot \left(\frac{1}{2}\right)^{10} = 1 \cdot \frac{1}{1024} = \frac{1}{1024} Total probability: 45+10+11024=561024\frac{45 + 10 + 1}{1024} = \frac{56}{1024}, which is answer C. Answer A (7128\frac{7}{128}) equals 561024\frac{56}{1024} when converted, but uses the wrong denominator initially. Answer B (11128\frac{11}{128}) likely comes from miscalculating the binomial coefficients. Answer D (1761024\frac{176}{1024}) suggests including cases with 6 or 7 correct answers, misinterpreting "more than 7." Remember: "more than 7" means 8, 9, or 10 — not 7 or more. Always identify exactly which outcomes satisfy the condition before calculating.